Answer:
a. 39.55%
b. 44.02%
Step-by-step explanation:
We have the following data:
n = 5
x = 1
p = 5/20 = 0.25
to. If the sampling is done with replacement.
We apply the binomial distribution formula, which is as follows:
P = nCx * (p ^ x) * ((1-p) ^ (n-x))
Where nCx, is a combination, and is equal to:
nCx = n! / x! * (n-x)!
replacing we have:
5C1 = 5! / 1! * 4! = 5
replacing in the main formula:
P = 5 * (0.25 ^ 1) * ((1- 0.25) ^ (5-1))
P = 0.3955
that is, without replacing the probability is 39.55%
b. if the sampling is done without replacement.
Here it is a little different from the previous one, but what you should do is calculate three cases,
the first was the one at point a, when n = 5 and x = 1
5C1 = 5! / 1! * 4! = 5
the second is when n = 20 and x = 5, this is all possible scenarios.
20C5 = 20! / 5! * 15! = 15504
and the third is when n = 15 (20-5) and x = 4 (5-1), which corresponds to the cases when none were damaged
15C4 = 15! / 4! * 11! = 1365
In the end, it would be:
P = (5C1 * 15C4) / 20C5
Replacing:
P = 5 * 1365/15504
P = 0.4402
Which means that without replacing the probability is 44.02%
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